Reaction Classes Precipitation: synthesis of an ionic solid a solid precipitate forms when aqueous solutions of certain ions are mixed AcidBase: proton transfer reactions acid donates a proton to a base, forming a molecule (water or another weak acid) and an aqueous salt Acid: protondonor; Base: protonacceptor OxidationReduction: electron transfer reactions electron transfer from one species to another, causing a change in the oxidation state of the two species OIL RIG: Oxidation Is Loss (of e ), Reduction Is Gain (of e ) includes combustion, the reaction of a substance with oxygen 1 Precipitation Reactions Sometimes when we mix two solutions together, an insoluble solid will form: AgNO 3 (aq) + NaCl(aq) AgCl(s) + NaNO 3 (aq) K 2 CrO 4 (aq) + Ba(NO 3 ) 2 (aq) BaCrO 4 (s) + 2 KNO 3 (aq) The solid, called a precipitate (or insoluble salt) is insoluble in water. It is so insoluble that when its component ions find each other in solution, they very rapidly get locked together in large clumps, driving the reaction towards the products. 2 1
The Solubility of Ionic Compounds in Water The solubility of ionic compounds in water depends upon the relative strengths of the electrostatic forces between ions in the ionic compound and the attractive forces between the ions and solvent molecules (often water). There is a tremendous range in the solubility of ionic compounds in water. The solubility of so called insoluble compounds may be several orders of magnitude less than ones that are called soluble in water. For example, consider the solubility (in g/l) of the following compounds in water at 20 o C : Solubility of NaCl = 365 Solubility of MgCl 2 = 542.5 Solubility of AlCl 3 = 699 Solubility of PbCl 2 = 9.9 Solubility of AgCl = 0.009 Solubility of CuCl = 0.0062 3 Precipitation Reactions The main challenge with precipitation reactions is predicting what solid (if any) will form. Example: Addition of potassium chromate to barium nitrate. K 2 CrO 4 (aq) + Ba(NO 3 ) 3 (aq) What is the precipitate? 4 2
Precipitation Reactions One approach: consider the ions present in the solutions: K 2 CrO 4 (aq) Ba(NO 3 ) 2 (aq) 5 Precipitation Reactions When combining ions, consider which precipitates could form Precipitates must have a net charge of zero Possible choices are BaCrO 4 and KNO 3 Clues: Potassium forms soluble salts. Chromate is yellow. BaCrO 4 precipitates! 6 3
Precipitation Reactions BaCl 2 (aq) AgNO 3 (aq) BaCl 2 (aq) + AgNO 3 (aq) Ba 2+ Cl Cl Ba 2+ Cl Cl NO 3 Ag + Ag + NO 3 Cl Ag + NO 3 Cl Cl Ag + Ba2+ NO 3 Cl Ba 2+ 2AgNO 3 (aq) + BaCl 2 (aq) 2AgCl(s) + Ba(NO 3 ) 2 (aq) Solid silver chloride precipitates out of solution. Barium and nitrate ions are left in aqueous solution. 7 Precipitation Reactions In the previous example, we know that Ba(NO 3 ) 2 is not the precipitate since in our first example this compound was soluble in water. Therefore, one way to determine if a precipitate will form is to simply study combinations of reactions where you know that one of the reaction products is soluble. That is exactly how precipitation or solubility rules have been determined 8 4
Solubility Rules Know these!!! In general, 14 are solubles and 56 are insolubles. 9 Precipitation Reactions: Will a Precipitate form? Example: If a solution containing potassium chloride is added to a solution containing ammonium nitrate, will a precipitate form? KCl(aq) + NH 4 NO 3 (aq) K + (aq) + Cl (aq) + NH 4+ (aq) + NO 3 (aq) Possible reaction products are KCl and NH 4 NO 3, or NH 4 Cl and KNO 3. All are soluble, so there is no precipitate. KCl(aq) + NH 4 NO 3 (aq) = No Reaction! Example: If a solution containing sodium iodide is added to a solution containing lead(ii) nitrate, will a precipitate form? NaI (aq) + Pb(NO 3 ) 2 (aq) Na + (aq) + I (aq) + Pb 2+ (aq) + 2 NO 3 (aq) Lead(II) iodide is insoluble; therefore a precipitate will form. 2 NaI (aq) + Pb(NO 3 ) 2 (aq) PbI 2 (s) + 2 NaNO 3 (aq) 10 5
Total and Net Ionic Equations Conventional (molecular) equation: a bookkeeping of all species present, and arranged for charge neutrality. Pb(NO 3 ) 2 (aq) + 2 NaI(aq) PbI 2 (s) + 2 NaNO 3 (aq) Total Ionic equation: all aqueous species are split up into their component ions. Pb 2+ (aq) + 2 NO 3 (aq) + 2 Na 2+ (aq) + 2 I (aq) Net Ionic equation: indicates exactly the chemical change that occurs, and nothing more. Pb 2+ (aq) + 2 I (aq) PbI 2 (s) + 2 Na + (aq) + 2 NO 3 (aq) PbI 2 (s) Na + and NO 3 are spectator ions in this reaction they are not involved in the chemical change. 11 Predicting Products of Reactions Write the complete ionic, net ionic, and conventional eqns. Al(NO 3 ) 3 (aq) + Ba(OH) 2 (aq) FeSO 4 (aq) + KCl(aq) CaCl 2 (aq) + Na 2 SO 4 (aq) Na 2 CrO 4 (aq) + AlBr 3 (aq) 6
Example Write the complete ionic, net ionic and conventional balanced chemical equations for the reaction of potassium sulfide and nickel(ii) nitrate. First: Remember nomenclature and write the ions in solution. K 2 S(aq) and Ni(NO 3 ) 2 (aq) gives: K + (aq) S 2 (aq) Ni 2+ (aq) NO 3 (aq) Next: Use the solubility rules to write the complete ionic eqn. 2 K + (aq) + S 2 (aq) + Ni 2+ (aq) + 2 NO 3 (aq) 2 K + (aq) + 2 NO 3 (aq) + NiS(s) Next: Cancel spectator ions on either side and write the net ionic eqn. Ni 2+ (aq) + S 2 (aq) NiS(s) Last: Write the conventional balanced equation. K 2 S(aq) + Ni(NO 3 ) 2 (aq) 2 KNO 3 (aq) + NiS (s) 13 We ve seen that some ionic compounds are soluble, while others are not. We can use this behavior to removes species selectively. Selective Precipitation Example: separating Ag + from Ba 2+ and Fe 3+. Notice that selective precipitation is nothing more than an application of the solubility rules. 14 7
Solubility and Stoichiometry I To determine the concentration of chloride ion in a sample of groundwater, a chemist adds 1.0 ml of 1.00 M AgNO 3 (aq) to 100.0 ml of the sample. The resulting mass of AgCl precipitate is 71.7 mg. What is the concentration of chloride in the original sample (expressed in mol Cl per L?) Ag + (aq) + Cl (aq) AgCl(s) Mass of AgCl Moles of AgCl Moles of Cl in 100.0 ml mol of Cl per liter 15 Example (cont) 1 g? mol Cl = 71.7 mg AgCl(s) 1000 mg = 0.00050028 mol Cl 1 mol AgCl 143.32 g AgCl 1 mol Cl 1 mol AgCl? mol/l Cl = 0.00050028 mol Cl 1000 ml 100.0 ml 1 L = 0.00500 M Cl = 5.00 x 10 3 M Cl 8
Solubility and Stoichiometry II What mass of solid NaCl must be added to 1.50 L of a 0.100 M AgNO 3 solution to precipitate all of the Ag +? Net Ionic Rxn: Ag + (aq) + Cl (aq) AgCl(s) First: Determine the moles of Ag+ in solution. 0.100 mol AgNO 1.50 L solution x 3 = 0.150 mol AgNO 1 L solution 3 1 mol Ag 0.150 mol AgNO 3 x + = 0.150 mol Ag 1 mol AgNO + 3 Then: Use stoichiometry of precipitation reaction to determine moles Cl needed, then the mass of NaCl. 0.150 mol Ag + 1 mol Cl x x 1 mol NaCl x 58.4 g NaCl = 8.76 g NaCl 1 mol Ag + 1 mol Cl 1 mol NaCl Solubility and Stoichiometry III We can also think about limiting reactants in precipitation reactions Calculate the mass of AgCl formed when 0.500 L of 1.20 M AgNO 3 is mixed with 0.250 L of 0.350 M NaCl. First: Determine the moles of Ag + and Cl. 1.20 mol AgNO 0.500 L solution x 3 = 0.600 mol AgNO 1 L solution 3 0.350 mol NaCl 0.250 L solution x = 0.0875 mol NaCl 1 L solution LR Then: Because it is a 1:1 rxn, the LR is easy to identify NaCl is the LR and the 0.0875 moles will dictate the mass of AgCl produced. 0.0875 mol Cl 1 mol AgCl x x 143.3 g AgCl = 12.5 g AgCl 1 mol Cl 1 mol AgCl 9
Reaction Classes Precipitation: synthesis of an ionic solid a solid precipitate forms when aqueous solutions of certain ions are mixed AcidBase: proton transfer reactions acid donates a proton to a base, forming a molecule (water or another weak acid) and an aqueous salt Acid: protondonor; Base: protonacceptor OxidationReduction: electron transfer reactions electron transfer from one species to another, causing a change in the oxidation state of the two species OIL RIG: Oxidation Is Loss (of e ), Reduction Is Gain (of e ) includes combustion, the reaction of a substance with oxygen 19 AcidBase Rxns BronstedLowry Theory: acid/base reactions are protontransfer processes. acid is protondonor (H + ion donor). base is proton acceptor (H + ion acceptor). B + H A B H + A When an acid gives its proton to water, water is acting as a base. H 2 O + H A H 2 O H + A When a base accepts a proton from water, water is acting as an acid. HO H + B HO + H B + 10
H 3 O + is called the hydronium ion Strong and Weak Acids Strong acids (think strong electrolyte ) undergo complete ionization. For HCl, essentially all the HCl molecules split up into ions. HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl (aq) Dominant species Weak acids (think weak electrolyte ) undergo incomplete ionization. For HF, only a very few H + and F ions exist in solution the reverse reaction dominates the chemistry Weak acids are like insoluble salts they don t like to dissociate very much. HF(aq) + H 2 O(l) H 3 O + (aq) + F (aq) Dominant species 11
Strong and Weak Acids Strong acids to know: Two wellknown oxyacids: nitric (HNO 3 ), sulfuric (H 2 SO 4 ) Group 7 acids: hydrochloric (HCl), hydrobromic (HBr), hydroiodic (HI) Two chlorine oxyacids: chloric (HClO 3 ), perchloric (HClO 4 ) Any other acids we see in this course will be weak. Polyprotic acids lose only their first H + easily, producing H 3 O + (aq) and an acid anion: H 2 SO 4 (aq) + H 2 O(l) HSO 4 (aq) + H 3 O + (aq) The acid anion of sulfuric acid, hydrogen sulfate anion, is also an acid, but it does not dissociate as readily as sulfuric acid it is a weak acid. Selected Acids and Bases Acids Bases Strong: H + (aq) + A (aq) Strong: M + (aq) + OH (aq) hydrochloric, HCl lithium hydroxide, LiOH hydrobromic, HBr sodium hydroxide, NaOH hydroiodoic, HI potassium hydroxide, KOH nitric acid, HNO 3 calcium hydroxide, Ca(OH) 2 sulfuric acid, H 2 SO 4 strontium hydroxide, Sr(OH) 2 perchloric acid, HClO 4 barium hydroxide, Ba(OH) 2 chloric acid, HClO 4 *(M is Group I or II metal) Weak Weak hydrofluoric, HF ammonia, NH 3 phosphoric acid, H 3 PO 4 accepts proton from water to make acetic acid, CH 3 COOH NH 4+ (aq) and OH (aq) (or HC 2 H 3 O 2 ) 12
Relative Acid Strength NOTE: The terms strong and weak refer only to how many acid (HA) molecules dissociate to H + and A. They have nothing to do with how corrosive the acid is. They also have nothing to do with concentration. Acid burn resulting from contact with HF, hyrdofluoric acid (a weak acid!). Predicting Products of Acid/Base Rxns Key to predicting the outcome of an acid/base reaction is identifying what is the acid and what is the base. 2 HClO 4 (aq) + Mg(OH) 2 (aq) 2 H 2 O(l) + Mg(ClO 4 ) 2 (aq) HCN(aq) + NaOH(aq) H 2 O(l) + NaCN(aq) CH 3 COOH(aq) + KOH(aq) H 2 O(l) + CH 3 COOK(aq) KF(aq) + H 2 O(l) HF(aq) + KOH(aq) H 2 PO 4 (aq) + 2 CH 3 COONa(aq) Na 2 HPO 4 (aq)+ 2 CH 3 COOH(aq) 13
Writing Balanced Equations for Neutralization Reactions I Write balanced descriptive equations (molecular, total ionic, and net ionic) for the following chemical reactions: a) calcium hydroxide(aq) and hydroiodic acid(aq) b) lithium hydroxide(aq) and nitric acid(aq) c) barium hydroxide(aq) and sulfuric acid(aq) These are all strong acids and bases, therefore, the products will be water and the corresponding salts. a) Ca(OH) 2 (aq) + 2HI (aq) CaI 2 (aq) + 2H 2 O (l) Ca 2+ (aq) + 2 OH (aq) + 2 H + (aq) + 2 I (aq) Ca 2+ (aq) + 2 I (aq) + 2 H 2 O (l) 2 OH (aq) + 2 H + (aq) 2 H 2 O (l) Writing Balanced Equations for Neutralization Reactions II b) LiOH (aq) + HNO 3 (aq) LiNO 3 (aq) + H 2 O (l) Li + (aq) + OH (aq) + H + (aq) + NO 3 (aq) Li + (aq) + NO 3 (aq) + H 2 O (l) OH (aq) + H + (aq) H 2 O (l) c) Ba(OH) 2 (aq) + H 2 SO 4 (aq) BaSO 4 (s) + 2 H 2 O (l) Ba 2+ (aq) + 2 OH (aq) + 2 H + (aq) + SO 4 2 (aq) BaSO 4 (s) + 2 H 2 O (l) Ba 2+ (aq) + 2 OH (aq) + 2 H + (aq) + SO 4 2 (aq) BaSO 4 (s) + 2 H 2 O (l) 14
Acid/Base and Stoichiometry What volume of 0.125 M HCl is needed to neutralize 200.0 ml of a 0.00955 M Ca(OH) 2 solution? Calculate the number of moles of base: V base x M base = 0.2000 L x 0.00955 M = 0.00191 mol Ca(OH) 2 From the balanced equation find the moles of acid needed: Ca(OH) 2(aq) + 2 HCl (aq) 2 H 2 O (l) + CaCl 2 (aq) Since there are two hydroxide ions per molecule of base and one proton per molecule of acid, we need twice as much acid as we have base: 0.00382 mol HCl Volume of acid: moles acid 0.00382 mol V acid = = = 0.0306 L HCl M acid 0.125 mol L Acid/Base and Limiting Reagent 75.0 ml of 0.250 M HCl is added to 225.0 ml of 0.0550 M Ba(OH) 2 solution. What is the concentration of the excess H + or OH ions left in this solution? 2 HCl(aq) + Ba(OH) 2 (aq) 2 H 2 O(l) + BaCl 2 (aq) 75.0 ml 225.0 ml 0.250 M 0.0550 M 1. How many moles of each reactant? 2. Which is limiting? 3. How much H + or OH is left over? 4. What is the new volume? 5. What is the concentration of the excess reactant? 15